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                Modern Art Plagiarism 树同构                </h1>
                <p>时间限制：3s&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  空间限制：128MB</p>			</div>
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                                <h3>题目描述</h3><p><p><font face="Times New Roman" size="3">定义树A与B同构，当且仅当树A的顶点经过一定的重新标号，树A的顶点集合边集完全与树B一一对应。 <br />
Your Task <br />
给定两棵树T1与T2，判断是否存在T1的子树T3，使得T3与T1同构。 <br />
</font></p></p><hr/><h3>输入格式</h3><p><p><font face="Times New Roman" size="3">第一行T表示数据组数 <br />
对于每组数据，第一行n表示T1的点数 <br />
接下来n-1行每行a b描述T1的一条边 <br />
接下来一行m表示T2的点数 <br />
再接下来m-1行每行a b描述T2的一条边 <br />
</font></p></p><hr/><h3>输出格式</h3><p><p><font face="Times New Roman" size="3">对于每组数据，在一行中输出YES表示存在T3，NO表示不存在 <br />
</font></p></p><hr/><h3>样例输入</h3><pre>51 2
2 3
3 4
4 5
4
1 2
1 3
1 4
5
1 2
1 3
1 4
4 5
4
1 2
2 3
3 4


</pre><hr/><h3>样例输出</h3><pre>	NO
	YES
</pre><hr/><h3>提示</h3><p><p>样例解释<br />
&nbsp;第一组数据中T2的点1度数为3，而T1是一条链，没有度数为3的点，显然NO。<br />
&nbsp;第二组数据中T2是一条4个点的链，与T1中的2-1-4-5同构。</p>
<p><br />
数据范围</p>
<p>&nbsp;100%：T&le;10，1&le;m&lt;=N&lt;=100<br />
</p></p><hr/><h3>题目来源</h3><p>google code jam 2008 APAC Onsites  </p>
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